Theorem: Let $f:\mathbb{R}\to\mathbb{C}$ be a function of rapid decay. Then $\sum_{n=-\infty}^\infty f(n)=\sum_{n=-\infty}^\infty \hat{f}(n)$, where $\hat{f}(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}\; dx$ is the Fourier transform of $f$.

Proof: Write $g(x)=\sum_{n=-\infty}^\infty f(x+n)$. The Fourier coefficients of $g$ are $a_n=\int_0^1\sum_{m=-\infty}^\infty f(x+m)e^{-2\pi inx}\; dx=\int_{-\infty}^\infty f(x)e^{-2\pi inx}\; dx=\hat{f}(n)$. Therefore the Fourier expansion of $g$ is $g(x)=\sum_{n=-\infty}^\infty \hat{f}(n)e^{2\pi inx}$. Substituting $x=0$ gives $\sum_{n=-\infty}^\infty f(n)=\sum_{n=-\infty}^\infty \hat{f}(n)$, as desired.